For the following code:
Two points to clarify:
1. Is the -v flag for declaring we're going to work with a variable?
2. Does the $LN stands for current line? If so, what's wrong with $0? Any difference between them?
Thanks
For the following code:
Two points to clarify:
1. Is the -v flag for declaring we're going to work with a variable?
2. Does the $LN stands for current line? If so, what's wrong with $0? Any difference between them?
Thanks
Hard to tell what you want - so I did this:
this is using AIX, different systems could handle this differently.
1) LN does not show to be a defined variable in my awk. (I even checked the gawk manual to see if it were there.) Why use $0 when just a simple print will default to the present line.
2) I was not sure if you wanted the current process ($$) or the parent process ($PPID} - ${PID}, was not defined in my version of AIX.
Hi "awk", Thanks and sorry for not clarifying enough.
The line is part of a running and working bash script on Linux.
PID is a local variable and the awk line is using its value to print the relevant line from ps.
I don't want to change the code (unless I have too), rather understand why it is used that way.
to simplify it, it is something like:
So, I understand the -v is used to declare variable, since ${MY_PID} will not be resolved inside 'single quotes' - right?
I still don't understand what is the $LN for?
Could anyone clarify? Is it a mistake, that works the same way as print $XXX works ???
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